The $n$th homology of convex subsets of Euclidean space using homotopic maps

Question

I confronted a problem when I was reading the theorem stating that if $f_0$ and $f_1$ are homotopic maps (not to be confused with chain homotopy notion) from $X$ to $Y$, then they induce the same natural homomorphism from the homology group of $X$ to that of $Y$.

Using this theorem, I want to prove that the $n$th homology group of a convex subset $U$ of $\mathbb R^n$ is zero. In fact the identity map on $U$ and the map $c$ on $U$ whose image is a single point $x\in U$ are homotopic. Now if I prove that the map $c$ induces the zero homomorphism on the $n$th homology group of $U$ then we are done. How can I show that?

Answer 1

You can view the map $c$ as a composite: $U \to \{x\} \to U$. Then the induced map on homology looks like $H_n(U) \to H_n(\{x\}) \to H_n(U)$. If you can show that $H_n(\{x\}=0$, then the induced map on homology must be the zero homomorphism.

Answer 2

If $U\subseteq \Bbb R^k$ is convex define $P_n:S_n(U)\rightarrow S_{n+1}(U)$ by $$P_n(\sigma)(\sum_{i=0}^{n+1}t_ie_i)=t_{n+1}x+(1-t_{n+1})\sigma(\sum_{i=0}^n\frac{t_i}{1-t_{n+1}}e_i)$$ for all singular simplexes $\sigma :\Delta^n\rightarrow U$, where $x\in U$. Then one can prove $$\partial_{n+1}(P_n(\sigma))=(-1)^{n+1}\sigma+P_{n-1}(\partial_n\sigma)$$, in particular which implies that $\sigma=(-1)^{n+1}\partial_{n+1}(P_n(\sigma))$ for all $\sigma\in Z_n(U)$ , for $n≥1$.